InsertName, That's an interesting formula! It's derivation is almost immediate, except I have doubts about the absence of the time coordinate, and the factors of 1/3.

Expand the metric in a Taylor's series:
gμν = Aμν + Bμνσxσ + Cμσντxσxτ + ...
It's always possible to choose coordinates such that Aμν = ημν and Bμνσ = 0, and in these coordinates the Christoffel symbols vanish. Then the formula for the Riemann tensor reduces to
Rμσντ = ½(gμτ,σν + gσν,μτ - gμν,στ - gστ,μν) = Cμστν + Cτνμσ - Cμσντ - Cσμτν.
If you assume C to have the same symmetry as the Riemann tensor, then this is 4Cμστν, showing that Cμστν = (1/4)Rμσντ